# Solution

## Transcription

Solution

Math 243 Spring 2015 (Practice) Midterm 2 4/2/2015 Time Limit: 50 minutes Name: • No calculators or notes are allowed. • One side of each sheet is blank and may be used as scratch paper. • Check your answers whenever possible. • Show your work clearly. Grade Table (for instructor use only) Question Points Score 1 15 2 20 3 20 4 15 5 30 Total: 100 Math 243 (Practice) Midterm 2 - Page 2 of 8 Common trig function values x sin x cos x tan x 0 0 1 0 π 6 1 2 π 4 √ 3 2 √1 3 √1 2 π 3 √1 2 √ 3 2 1 2 1 √ 3 π 2 1 0 undefined Trig identities • sin2 x + cos2 x = 1 • tan2 x + 1 = sec2 x • sin2 x = 1−cos 2x 2 • cos2 x = 1+cos 2x 2 • sin 2x = 2 sin x cos x 4/2/2015 The cross product If ~u = ha, b, ci and ~v = hp, q, ri, then ~u × ~v = hbr − cq, cp − ar, aq − bpi. Differential geometry d~r ~v • Tˆ = = ds |~v | ˆ ˆ ˆ = 1 dT = 1 dT • N dTˆ dt κ ds dt dTˆ dTˆ 1 • κ= = ds |~v | dt Z b |~v (t)| dt • arclength = a Math 243 (Practice) Midterm 2 - Page 3 of 8 4/2/2015 1. (15 points) Find an equation for the plane that is parallel to the vectors h2, 1, −1i and h0, −2, 3i and that passes through the point (1, 2, 3). Solution: First take the cross product of the two vectors to find a vector that is normal to the plane: h2, 1, −1i × h0, −2, 3i = h1, −6, −4i. The equation for the plane is: x − 6y − 4z = h1, −6, −4i · h1, 2, 3i = 23. Math 243 (Practice) Midterm 2 - Page 4 of 8 4/2/2015 2. Consider the planes 3x + 6z = 1 and 2x + 2y − z = 3. (a) (10 points) What is the angle between these planes? Solution: We have to find the angle between the normal vectors h3, 0, 6i and h2, 2, −1i. The dot product of these two vectors is 0, so they are orthogonal. The planes are at a right angle to each other. (b) (10 points) Find a parametrization for the line of intersection of these planes. Solution: First, notice that the point (1/3, 7/6, 0) is on the line (set z = 0). The cross product of the two normal vectors, h−12, 15, 6i is parallel to the line. So we have ~r(t) = h1/3, 7/6, 0i + th−12, 15, 6i. Math 243 (Practice) Midterm 2 - Page 5 of 8 4/2/2015 3. A particle moves with constant acceleration ~a = h0, 0, −1i, starting at the point (0, 0, 1) and with an initial velocity of h1, 1, 0i. (a) (10 points) When and where does the particle fall below the plane z = 0? Solution: First, ~v (t) = h1, 1, −ti, so t2 . ~r(t) = t, t, 1 − 2 √ 2 The z coordinate is 0 when t2 = 1, i.e., when t = 2. (b) (10 points) Find a vector equation for the tangent line to the particle’s trajectory at the point where it hits the plane z = 0. √ √ √ √ √ Solution: ~v ( 2) = h1, 1, − 2i, and ~ r ( 2) = h √ √ √ 2, 2, 0i, so the equation of ~ the tangent line is l(s) = h 2, 2, 0i + sh1, 1, − 2i. Math 243 (Practice) Midterm 2 - Page 6 of 8 4/2/2015 4. (15 points) A particle’s position at time t is given by ~r(t) = h1, 3t2 , t3 i. Find the length of the arc traced out by the particle between time t = 0 and t = 1. Solution: Z 1 Z 1 |~v (t)| dt = 0 |~v (t)| dt 0 Z 1 = √ 36t2 + 9t4 dt 0 Z = 1 √ 3t 4 + t2 dt 0 1 = (4 + t2 )3/2 0 = 53/2 − 8 Math 243 (Practice) Midterm 2 - Page 7 of 8 4/2/2015 5. A particle’s position at time t is given by ~r(t) = h1, et cos t − 1, et sin ti. (a) (10 points) Find the particle’s speed at time t. Solution: ~v (t) = h0, et cos t − et sin t, et sin t + et cos ti = et h0, cos t − sin t, cos t + sin ti. So the speed is |~v (t)| = √ 2et . (b) (10 points) Find Tˆ(t), the unit tangent vector to the particle’s trajectory at time t. Solution: ~v (t) 1 Tˆ(t) = = √ h0, cos t − sin t, cos t + sin ti. |~v (t)| 2 Math 243 (Practice) Midterm 2 - Page 8 of 8 4/2/2015 (c) (5 points) Find the curvature κ(t). Solution: 1 dTˆ 1 κ(t) = = √ t |~v (t)| dt 2e 1 √ h0, − sin t − cos t, − sin t + cos ti = √1 . 2 2et ˆ (t). (d) (5 points) Find the unit normal vector N Solution: ˆ (t) = √1 h0, − sin t − cos t, − sin t + cos ti. N 2